Giải hệ phương trình : {8 /(x-3) + 1/( 2|y|-3 ) =5 {4/( x-3) +1/( 2|y|-3)=3

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1. Đáp án:

   \(\left\{ \begin{array}{l}
   x = 5\\
   \left[ \begin{array}{l}
   y = 2\\
   y =  – 2
   \end{array} \right.
   \end{array} \right.\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   DK:x \ne 3;y \ne  \pm \frac{3}{2}\\
   \left\{ \begin{array}{l}
   \frac{8}{{x – 3}} + \frac{1}{{2\left| y \right| – 3}} = 5\\
   \frac{4}{{x – 3}} + \frac{1}{{2\left| y \right| – 3}} = 3
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   \frac{8}{{x – 3}} + \frac{1}{{2\left| y \right| – 3}} = 5\\
    – \frac{4}{{x – 3}} – \frac{1}{{2\left| y \right| – 3}} =  – 3
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   \frac{{8 – 4}}{{x – 3}} = 5 – 3\\
   \frac{8}{{x – 3}} + \frac{1}{{2\left| y \right| – 3}} = 5
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   \frac{4}{{x – 3}} = 2\\
   \frac{8}{{x – 3}} + \frac{1}{{2\left| y \right| – 3}} = 5
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x – 3 = 2\\
   \frac{8}{{x – 3}} + \frac{1}{{2\left| y \right| – 3}} = 5
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x = 5\\
   \frac{8}{{5 – 3}} + \frac{1}{{2\left| y \right| – 3}} = 5
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x = 5\\
   \frac{1}{{2\left| y \right| – 3}} = 5 – 4 = 1
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x = 5\\
   2\left| y \right| – 3 = 1
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x = 5\\
   2\left| y \right| = 4
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x = 5\\
   \left[ \begin{array}{l}
   y = 2\\
   y =  – 2
   \end{array} \right.
   \end{array} \right.\left( {TM} \right)
   \end{array}\)

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